Time dependent perturbation for harmonic oscillator

[Vandmelon]

Homework Statement

I'm looking at the 1d harmonic oscillator
\begin{equation}
V(x)=\frac{1}{2}kx^2
\end{equation}
with eigenstates n and the time dependent perturbation
\begin{equation}
H'(t)=qx^3\frac{(\tau^2}{t^2+\tau^2}
\end{equation}
For t=-∞ the oscillator is in the groundstate n=0
I need to show that for n>3 the states will not get excited and I only need to look at the first order perturbation.

The Attempt at a Solution

What I'm thinking is \begin{equation}P_{a→b}=\vert c_b(t)\vert^2 \end{equation} therefore I need to find when c_b=0
and because
\begin{equation}
c_b'=-\frac{i}{\hbar}∫H_{ba}'(t')e^{i\omega_0t'}dt'
\end{equation}
I need to find
\begin{equation}
H_{ba}'=<\psi_n\vert H' \vert \psi_a >=0
\end{equation}
For n>3
It would make sense if H_ba look a bit like this
\begin{equation}
H_{ba}'=\frac{d^n}{dx^n}H'
\end{equation}
But I get some horrible results if I try to solve H_ba. Is it right what I am doing or is it way off.

 

[fzero]

You want to compute $\langle n| H' |0\rangle$, which is proportional to $\langle n| \hat{x}^3 |0\rangle$. If you express $\hat{x}$ in terms of raising and lowering operators $\hat{a},\hat{a}^\dagger$, you should be able to see how to reach the necessary result.

 

[Vandmelon]

so if I use
\begin{equation}
\hat x=\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)
\end{equation}

then if I solved it right
\begin{equation}
\hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3+a_+ + a_- + (a_-)^3]\psi_0
\end{equation}
Is it correct if I use a_- on ψ_0 it will just be 0 so
\begin{equation}
\hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3\psi_0+a_+\psi_0]=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[\sqrt6 \psi_3+\psi_1]
\end{equation}
So it will end up like this
\begin{equation}
\langle \psi_n \vert\sqrt6 \psi_3+\psi_1 \rangle
\end{equation}
If this is correct I am not sure why the oscillator wouldn't get excited for n>3. Or is it my calculations there are wrong?

*EDIT
Oh is it because
\begin{equation}
\langle \psi_n \vert \psi_m \rangle = 0
\end{equation}
for n ≠ m

 

[fzero]

so if I use
\begin{equation}
\hat x=\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)
\end{equation}

then if I solved it right
\begin{equation}
\hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3+a_+ + a_- + (a_-)^3]\psi_0
\end{equation}

You'll want to go back through the algebra here. Remember that $a_+,a_-$ don't commute, so you have to keep track of the order of terms.

Is it correct if I use a_- on ψ_0 it will just be 0 so

...

Oh is it because
\begin{equation}
\langle \psi_n \vert \psi_m \rangle = 0
\end{equation}
for n ≠ m

You will have to use both of these relations. You have all of the right ideas, just fix the algebra.

 

[paranormal]

I can provide some guidance on approaching this problem. First, it is important to note that the time dependent perturbation Hamiltonian, H'(t), is only relevant for times greater than or equal to 0. This is because the perturbation is described as occurring at t=0. Therefore, for t=-∞, the system is in the ground state and the perturbation has not yet occurred.

Now, to determine when the states will not get excited, we can use the first order perturbation theory. This means that we can ignore higher order terms in the perturbation, such as the term involving x^3. So, we only need to consider the first order term in the perturbation Hamiltonian, which is proportional to x.

To find when c_b=0, we can use the time evolution of the coefficients, c_b'. As you have correctly stated, c_b' is given by the integral involving H_ba'. However, in this case, H_ba' is not equal to the nth derivative of H', as you have suggested. Instead, it is given by the matrix element <ψ_n|H'|ψ_a>, where ψ_n and ψ_a are the eigenstates of the unperturbed Hamiltonian, H_0. In this case, H_0 is the harmonic oscillator Hamiltonian, which has eigenstates n. Therefore, we need to calculate the matrix element <n|H'|m>, where n and m are integers representing the eigenstates. This can be done using the harmonic oscillator wavefunctions and the corresponding energy eigenvalues.

Once we have the expression for H_ba', we can then substitute it into the integral for c_b' and solve for the conditions when c_b=0. This will give us the values of n and m for which the states will not get excited.

In summary, to show that for n>3 the states will not get excited, we need to use first order perturbation theory and calculate the matrix element <n|H'|m>. Then, we can use this expression to solve for when c_b=0 and determine the values of n and m for which the states will not get excited.